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# If a wire is bent into the shape of a square, then the area enclosed

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If a wire is bent into the shape of a square, then the area enclosed  [#permalink]

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31 Jan 2019, 23:45
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Difficulty:

85% (hard)

Question Stats:

46% (02:54) correct 54% (02:41) wrong based on 48 sessions

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If a wire is bent into the shape of a square, then the area enclosed by the square is 81 cm square. When the same wire is bent into semi-circle. then the area of semi circle will be

A. 22 cm square
B. 44 cm square
C. 77 cm square
D. 81 cm square
E. 154 cm square

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Re: If a wire is bent into the shape of a square, then the area enclosed  [#permalink]

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01 Feb 2019, 01:00
Bunuel wrote:
If a wire is bent into the shape of a square, then the area enclosed by the square is 81 cm square. When the same wire is bent into semi-circle. then the area of semi circle will be

A. 22 cm square
B. 44 cm square
C. 77 cm square
D. 81 cm square
E. 154 cm square

Length of the wire=Perimeter of the square=$$4*\sqrt{81}$$=4*9=36

When the same wire is bent into semi-circle, then the perimeter of the semi-circle=36
Or, $$\pi*r+2r=36$$
Or, $$r(\pi+2)=36$$
Or, $$r=7$$cm

Now area of semi-circle=$$\frac{1}{2}*\pi*7^2=77^2$$

Ans. (C)
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Re: If a wire is bent into the shape of a square, then the area enclosed  [#permalink]

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02 Feb 2019, 00:53
Bunuel wrote:
If a wire is bent into the shape of a square, then the area enclosed by the square is 81 cm square. When the same wire is bent into semi-circle. then the area of semi circle will be

A. 22 cm square
B. 44 cm square
C. 77 cm square
D. 81 cm square
E. 154 cm square

given area of square = s^2 = 81= s=9
perimter of square = 4*s= 36
so
perimeter of semi circle = 36
pi * r+2r = 36
r = 7
area of semi circle = pi * r^2 /2
= 77
IMO C
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Re: If a wire is bent into the shape of a square, then the area enclosed  [#permalink]

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05 Feb 2019, 04:40
Area of square= 81
Side of square= 9
Perimeter of square = length of wire= 36
Circumference of semicircle= pi.r + 2r = r(pi+2) = 36
Area of semicircle= pi.r^2/2= pi.36^2/2.(pi+2)^2= pi*36*18/(5.16)^2= (3.16*2*18*18)/(5.16*5.16)
Now, 18/5.16 is approx 3.5 (half way between 5.16*3=15.48 and 5.16*4=20.64)
So, approx area of semicircle= 3.16*2*3.5*3.5= (3.16*2*49)/4= (3.16*49)/2= 155/2= 77 (approx)
Ans C

Caution: Approx. calculations are effective only when ans choices are sufficiently apart.
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Posts: 4
Re: If a wire is bent into the shape of a square, then the area enclosed  [#permalink]

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05 Feb 2019, 05:04
1
Given area=81
side of square= 9
perimeter =4*9=36
But same wire is turned into semi circle so
pi.r+2*r=36
(22/7 +2)*r=36
36/7 *r =36
area of semi circle = (1/2)*(22/7)*7*7=77
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Re: If a wire is bent into the shape of a square, then the area enclosed  [#permalink]

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06 Feb 2019, 19:45
Bunuel wrote:
If a wire is bent into the shape of a square, then the area enclosed by the square is 81 cm square. When the same wire is bent into semi-circle. then the area of semi circle will be

A. 22 cm square
B. 44 cm square
C. 77 cm square
D. 81 cm square
E. 154 cm square

Since the area of the square is 81, the side is 9, and thus the perimeter is 36. Since 36 is the perimeter of the semicircle, we know that, in terms of the circle’s radius, the semicircle’s perimeter is twice the radius and half the circumference (of the entire circle). Thus, we can create the following equation in which r = radius:

2r + 2πr/2 = 36

2r + πr = 36

r(2 + π) = 36

r = 36/(2 + π)

Since π is slightly greater than 3, r is approximately 36 divided by a number slightly greater than 5. So we can approximate r as 7. Using this approximation, we calculate the area of the semicircle as:

1/2 x π x 7^2

1/2 x π x 49

If we round π down to 3 and 49 up to 50, we have:

1/2 x 3 x 50 = 75

So the area of the semicircle is approximately 75 square cm. We see that choice C must be the correct answer.

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Re: If a wire is bent into the shape of a square, then the area enclosed   [#permalink] 06 Feb 2019, 19:45
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